Class 12: Inverse Trigonometric Functions – Exercise 4.14 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1: Evaluate the following:}$

$\displaystyle \text{(i) } \tan \Bigg\{ 2 \tan^{-1} \frac{1}{5} - \frac{\pi}{4} \Bigg\}$

$\displaystyle \text{(ii) } \tan \Bigg\{ \frac{1}{2} \sin^{-1} \frac{3}{4} \Bigg\} {\hspace{5.0cm} \text{[CBSE 2013]} }$

$\displaystyle \text{(iii) } \sin \Bigg\{ \frac{1}{2} \cos^{-1} \frac{4}{5} \Bigg\}$

$\displaystyle \text{(iv) } \sin \Bigg\{ 2 \tan^{-1} \frac{2}{3} \Bigg\} + \cos \Bigg\{ \tan^{-1} \sqrt{3} \Bigg\}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$

$\displaystyle \tan \Bigg\{ 2 \tan^{-1} \frac{1}{5} - \frac{\pi}{4} \Bigg\}$

$\displaystyle = \tan \Bigg\{ 2 \tan^{-1} \frac{1}{5} - \tan^{-1} 1 \Bigg\}$

$\displaystyle = \tan \Bigg\{ \tan^{-1} \Big\{ \frac{\frac{1}{5} +\frac{1}{5} }{1 - \frac{1}{5} \times \frac{1}{5} } \Big\} - \tan^{-1} 1 \Bigg\}$

$\displaystyle = \tan \Bigg\{ \tan^{-1} \Big\{ \frac{\frac{2}{5} }{\frac{24}{25} } \Big\} - \tan^{-1} 1 \Bigg\}$

$\displaystyle = \tan \Bigg\{ \tan^{-1} \frac{5}{12} - \tan^{-1} 1 \Bigg\}$

$\displaystyle \Big[ \because \tan^{-1} x - \tan^{-1} y = \tan^{-1} \Big( \frac{x+y}{1-xy} \Big) \Big]$

$\displaystyle = \tan \Bigg\{ \frac{\frac{5}{12} -1 }{1 + \frac{5}{12} \times 1 } \Bigg\}$

$\displaystyle = \tan \Bigg\{ \tan^{-1} \frac{-\frac{7}{12}}{\frac{17}{12}} \Bigg\}$

$\displaystyle = \tan \Bigg\{ \tan^{-1} \frac{-7}{12} \Bigg\}$

$\displaystyle = \frac{-7}{12}$

$\displaystyle \text{(ii)}$

$\displaystyle \text{Given, } \tan \Bigg\{ \frac{1}{2} \sin^{-1} \frac{3}{4} \Bigg\}$

$\displaystyle \text{Let, } \frac{1}{2} \sin^{-1} \frac{3}{4} = x$

$\displaystyle \Rightarrow \sin^{-1} \frac{3}{4} = 2x$

$\displaystyle \Rightarrow \sin 2x = \frac{3}{4}$

$\displaystyle \Rightarrow \cos 2x = \frac{\sqrt{7}}{4}$

$\displaystyle \Rightarrow \frac{1}{2} \sin^{-1} \frac{3}{4} = \tan x = \sqrt{ \frac{1 - \cos 2x }{1+ \cos 2x} } = \sqrt{ \frac{4 - \sqrt{7}}{4 - \sqrt{7}} } = \sqrt{ \frac{(4-\sqrt{7})^2}{9} } = \frac{4 - \sqrt{7}}{3}$

$\displaystyle \text{(iii)}$

$\displaystyle\sin \Bigg\{ \frac{1}{2} \cos^{-1} \frac{4}{5} \Bigg\}\ \ \ \ \ \Bigg[ \because \cos^{-1}x = 2 \sin^{-1} \pm \sqrt{ \frac{1 - x}{2} } \Bigg]$

$\displaystyle= \sin \Bigg\{ \frac{1}{2} \sin^{-1} \pm \sqrt{ \frac{1 - \frac{4}{5}}{2} } \Bigg\}$

$\displaystyle= \pm 10$

$\displaystyle \text{(iv)}$

$\displaystyle\sin \Bigg\{ 2 \tan^{-1} \frac{2}{3} \Bigg\} + \cos \Bigg\{ \tan^{-1} \sqrt{3} \Bigg\}$

$\displaystyle= \sin \Bigg\{ 2 \sin^{-1} \frac{2 \times \frac{2}{3}}{1 + \frac{4}{9}} \Bigg\} + \cos \Bigg\{ \cos^{-1} \frac{1}{\sqrt{1 + (\sqrt{3})^2}} \Bigg\}$

$\displaystyle= \sin \Bigg( \sin^{-1} \frac{12}{13} \Bigg) + \cos \Bigg( \cos^{-1} \frac{1}{2} \Bigg)$

$\displaystyle= \frac{12}{13}+ \frac{1}{2}$

$\displaystyle= \frac{37}{36}$

$\\$

$\displaystyle \text{Question 2: Prove the following results:}$

$\displaystyle \text{(i) } 2 \sin^{-1} \frac{3}{5} = \tan^{-1}\frac{24}{7}$

$\displaystyle \text{(ii) } \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9} = \frac{1}{2} \cos^{-1} \frac{3}{5} = \frac{1}{2} \sin^{-1} \frac{4}{5} {\hspace{5.0cm} \text{[CBSE 2010]} }$

$\displaystyle \text{(iii) } \tan^{-1} \frac{2}{3} = \frac{1}{2} \tan^{-1}\frac{12}{5}$

$\displaystyle \text{(iv) } \tan^{-1} \frac{1}{7} + 2\tan^{-1} \frac{1}{3} = \frac{\pi}{4} {\hspace{7.0cm} \text{[CBSE 2010]} }$

$\displaystyle \text{(v) } \sin^{-1} \frac{4}{5} + 2\tan^{-1} \frac{1}{3} = \frac{\pi}{2}$

$\displaystyle \text{(vi) } 2\sin^{-1} \frac{3}{5} - \tan^{-1} \frac{17}{31} = \frac{\pi}{4}$

$\displaystyle \text{(vii) } 2\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{8} = \tan^{-1} \frac{4}{7}$

$\displaystyle \text{(viii) } 2\tan^{-1} \frac{3}{4} - \tan^{-1} \frac{17}{31} = \frac{\pi}{4} {\hspace{7.0cm} \text{[CBSE 2011]} }$

$\displaystyle \text{(ix) } 2\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{7} = \tan^{-1} \frac{17}{31} {\hspace{6.0cm} \text{[CBSE 2011]} }$

$\displaystyle \text{(x) } 4\tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239} = \frac{\pi}{4}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$

$\displaystyle \text{LHS } = 2 \sin^{-1} \frac{3}{5} \ \ \ \ \ \Bigg[ \because \sin^{-1} x = \tan^{-1} \Big(\frac{x}{\sqrt{1 - x^2}} \Big)\Bigg]$

$\displaystyle = 2 \tan^{-1} \Bigg[ \frac{\frac{3}{4}}{\sqrt{1 - \frac{9}{25}}} \Bigg]$

$\displaystyle = 2 \tan^{-1} \Bigg[ \frac{\frac{3}{5}}{\frac{4}{5}} \Bigg]$

$\displaystyle = 2 \tan^{-1} \frac{3}{4} \ \ \ \ \ \Bigg[ \because 2\tan^{-1} x = \tan^{-1} \Big(\frac{2x}{1 - x^2} \Big) \Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \frac{2 \times \frac{3}{4} }{1 - (\frac{3}{4})^2}\Bigg]$

$\displaystyle = \tan^{-1} \Bigg[ \frac{\frac{3}{2}}{\frac{7}{16}}\Bigg]$

$\displaystyle = \tan^{-1} \frac{24}{7} = \text{ RHS }$

$\displaystyle \text{(ii)}$

$\displaystyle\tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2}{9}$

$\displaystyle= \tan^{-1} \Bigg( \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4} \cdot \frac{2}{9}} \Bigg)$

$\displaystyle= \tan^{-1} \Bigg( \frac{\frac{17}{36}}{\frac{34}{36}} \Bigg)$

$\displaystyle= \tan^{-1} \frac{1}{2}$

$\displaystyle= \frac{1}{2} \cos^{-1} \Bigg( \frac{1- \frac{1}{4}}{1+ \frac{1}{4}} \Bigg)$

$\displaystyle= \frac{1}{2} \cos^{-1} \Bigg( \frac{\frac{3}{4}}{\frac{5}{4}} \Bigg)$

$\displaystyle= \frac{1}{2} \cos^{-1} \frac{3}{5}$

$\displaystyle \text{Now, } \tan^{-1} \frac{1}{2} = \frac{1}{2} \sin^{-1} \Bigg( \frac{\frac{2}{2}}{1 + \frac{1}{4}} \Bigg)$

$\displaystyle= \frac{1}{2} \sin^{-1} \Bigg( \frac{\frac{1}{5}}{\frac{5}{4}} \Bigg)$

$\displaystyle= \frac{1}{2} \sin^{-1} \frac{4}{5}$

$\displaystyle \text{(iii)}$

$\displaystyle \text{LHS } = \tan^{-1} \frac{2}{3}$

$\displaystyle = \frac{1}{2} \tan^{-1} \Bigg\{ \frac{2 \times \frac{2}{3} }{1 - ( \frac{2}{3})^2} \Bigg\}$

$\displaystyle = \frac{1}{2} \tan^{-1} \Bigg\{ \frac{\frac{4}{3}}{\frac{5}{9}} \Bigg\}$

$\displaystyle = \frac{1}{2} \tan^{-1}\frac{12}{5} = \text{RHS }$

$\displaystyle \text{(iv)}$

$\displaystyle \text{LHS } = \tan^{-1} \frac{1}{7} + 2\tan^{-1} \frac{1}{3}$

$\displaystyle = \tan^{-1} \frac{1}{7} + \tan^{-1} \Bigg\{ \frac{2 \times \frac{1}{3} }{1 - ( \frac{1}{3})^2} \Bigg\}$

$\displaystyle = \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{3}{4}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{1}{7}+\frac{3}{4}}{1 -\frac{1}{7} \cdot\frac{3}{4} } \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{25}{28}}{\frac{25}{28}} \Bigg\}$

$\displaystyle = \tan^{-1} 1 = \frac{\pi}{4} = \text{RHS }$

$\displaystyle \text{(v)}$

$\displaystyle \text{LHS } = \sin^{-1} \frac{4}{5} + 2\tan^{-1} \frac{1}{3}$

$\displaystyle = \sin^{-1} \frac{4}{5} + \tan^{-1} \Bigg\{ \frac{2 \times \frac{1}{3} }{1 - ( \frac{1}{3})^2} \Bigg\}$

$\displaystyle = \sin^{-1} \frac{4}{5} + \tan^{-1} \Bigg\{ \frac{\frac{2}{3}}{\frac{8}{9}} \Bigg\}$

$\displaystyle = \sin^{-1} \frac{4}{5} + \tan^{-1} \frac{3}{4}$

$\displaystyle = \sin^{-1} \frac{4}{5} + \cos^{-1} \Bigg\{ \frac{1}{\sqrt{1 + \frac{9}{16}}} \Bigg\}$

$\displaystyle = \sin^{-1} \frac{4}{5} + \cos^{-1} \frac{4}{5}$

$\displaystyle = \frac{\pi}{2} = \text{RHS }$

$\displaystyle \text{(vi)}$

$\displaystyle \text{LHS } = 2\sin^{-1} \frac{3}{5} - \tan^{-1} \frac{17}{31}$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{\frac{3}{4}}{\sqrt{1 - \frac{9}{25}}} \Bigg\}- \tan^{-1} \frac{17}{31}$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{\frac{3}{5}}{\frac{4}{5}} \Bigg\}- \tan^{-1} \frac{17}{31}$

$\displaystyle = 2\tan^{-1} \frac{3}{4}- \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2 \times \frac{3}{4} }{1 - ( \frac{3}{4})^2} \Bigg\} - \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{3}{2}}{\frac{7}{16}} \Bigg\}- \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{24}{7} \Bigg\}- \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{24}{7} - \frac{17}{31}}{1 + \frac{24}{7} \cdot \frac{17}{31}} \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{625}{217}}{\frac{625}{217}} \Bigg\}$

$\displaystyle = \tan^{-1} 1 = \frac{\pi}{4} = \text{RHS }$

$\displaystyle \text{(vii)}$

$\displaystyle \text{LHS } = 2\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{8}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2 \times \frac{1}{5} }{1 - ( \frac{1}{5})^2} \Bigg\} + \tan^{-1} \frac{1}{8}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{2}{5}}{\frac{24}{25}} \Bigg\}+ \tan^{-1} \frac{1}{8}$

$\displaystyle = \tan^{-1} \frac{5}{12}+ \tan^{-1} \frac{1}{8}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{5}{12}+\frac{1}{8}}{1 -\frac{5}{12} \cdot\frac{1}{8} } \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{13}{24}}{\frac{91}{96}} \Bigg\}$

$\displaystyle = \tan^{-1} \frac{4}{7} = \text{RHS }$

$\displaystyle \text{(viii)}$

$\displaystyle \text{LHS } = 2\tan^{-1} \frac{3}{4} - \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2 \times \frac{3}{4} }{1 - ( \frac{3}{4})^2} \Bigg\} - \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{3}{2}}{\frac{7}{16}} \Bigg\} - \tan^{-1} \frac{17}{31}$

$\displaystyle = \tan^{-1} \frac{24}{7} - \tan^{-1} \frac{17}{31}$

$\displaystyle = \Bigg\{ \frac{\frac{24}{7}-\frac{17}{31}}{1 +\frac{24}{7} \cdot\frac{17}{31} } \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{625}{217}}{\frac{625}{217}} \Bigg\}$

$\displaystyle = \tan^{-1} 1 = \frac{\pi}{4} = \text{RHS }$

$\displaystyle \text{(ix)}$

$\displaystyle \text{LHS } = 2\tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{7}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2 \times \frac{1}{2} }{1 - ( \frac{1}{2})^2} \Bigg\} + \tan^{-1} \frac{1}{7}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{1}{\frac{3}{4}} \Bigg\} + \tan^{-1} \frac{1}{7}$

$\displaystyle = \tan^{-1} \Big\{ \frac{4}{3} \Big\} + \tan^{-1} \frac{1}{7}$

$\displaystyle = \tan^{-1}\Bigg\{ \frac{\frac{4}{3}+\frac{1}{7}}{1 -\frac{4}{3} \cdot\frac{1}{7} } \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{31}{21}}{\frac{17}{21}} \Bigg\}$

$\displaystyle = \tan^{-1} \frac{31}{17} = \text{RHS }$

$\displaystyle \text{(ix)}$

$\displaystyle \text{LHS } = 4\tan^{-1} \frac{1}{5} - \tan^{-1} \frac{1}{239}$

$\displaystyle = 2 ( 2\tan^{-1} \frac{1}{5} ) - \tan^{-1} \frac{1}{239}$

$\displaystyle = 2 ( \tan^{-1} \Bigg\{ \frac{2 \times \frac{1}{5} }{1 - ( \frac{1}{5})^2} \Bigg\} ) - \tan^{-1} \frac{1}{239}$

$\displaystyle = 2 (\tan^{-1} \Bigg\{ \frac{\frac{2}{5}}{\frac{24}{25}} \Bigg\} ) - \tan^{-1} \frac{1}{239}$

$\displaystyle = 2 (\tan^{-1} \Bigg\{ \frac{5}{12} \Bigg\} ) - \tan^{-1} \frac{1}{239}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2 \times \frac{5}{12} }{1 - ( \frac{5}{12})^2} \Bigg\} - \tan^{-1} \frac{1}{239}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{5}{6}}{\frac{119}{144}} \Bigg\} - \tan^{-1} \frac{1}{239}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{120}{119} \Bigg\} - \tan^{-1} \frac{1}{239}$

$\displaystyle = \tan^{-1}\Bigg\{ \frac{\frac{120}{119}-\frac{1}{239}}{1 +\frac{120}{119} \cdot\frac{1}{239} } \Bigg\}$

$\displaystyle = \tan^{-1} 1 = \frac{\pi}{4} = \text{RHS }$

$\\$

$\displaystyle \text{Question 3: If: } \sin^{-1} \frac{2a}{1+a^2} - \cos^{-1} \frac{1-b^2}{1+b^2} = \tan^{-1} \frac{2x}{1-x^2}. \text{ then prove that } \\ x = \frac{a-b}{1+ab}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } a = \tan m, \ \ \ b = \tan n \ \ \ x = \tan y$

$\displaystyle \text{Now, } \sin^{-1} \frac{2a}{1+a^2} - \cos^{-1} \frac{1-b^2}{1+b^2} = \tan^{-1} \frac{2x}{1-x^2}$

$\displaystyle \Rightarrow \sin^{-1} \frac{2 \tan m}{1+\tan^2 m} - \cos^{-1} \frac{1-\tan n^2}{1+\tan^2 n} = \tan^{-1} \frac{2 \tan y}{1-\tan^2 y}$

$\displaystyle \Rightarrow \sin^{-1} (\sin 2m) - \cos^{-1} (\cos 2n) = \tan^{-1} ( \tan^2 y)$

$\displaystyle \Rightarrow 2m - 2n = 2y$

$\displaystyle \Rightarrow m - n = y$

$\displaystyle \Rightarrow \tan^{-1} a - \tan^{-1} b= \tan^{-1} x$

$\displaystyle \Rightarrow \tan^{-1} \frac{a-b}{1+ab} = \tan^{-1} x$

$\displaystyle \Rightarrow \frac{a-b}{1+ab} = x$

$\\$

$\displaystyle \text{Question 4: Prove that:}$

$\displaystyle \text{(i)} \tan^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg) + \cot^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg)= \frac{\pi}{2}$

$\displaystyle \text{(ii)} \sin \Bigg\{ \tan^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg) + \cos^{-1} \Bigg( \frac{1-x^2}{1+x^2} \Bigg) \Bigg\} = 1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$

$\displaystyle \text{LHS } = \tan^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg) + \cot^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg)$

$\displaystyle = \tan^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg) + \frac{\pi}{2} - \tan^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg)$

$\displaystyle = \frac{\pi}{2} = \text{RHS }$

$\displaystyle \text{(ii)}$

$\displaystyle \text{LHS } = \sin \Bigg\{ \tan^{-1} \Bigg( \frac{1-x^2}{2x} \Bigg) + \cos^{-1} \Bigg( \frac{1-x^2}{1+x^2} \Bigg) \Bigg\}$

$\displaystyle = \sin \Bigg\{ \sin^{-1} \Bigg( \frac{\frac{1-x^2}{2x}}{\sqrt{1+ \frac{1-x^2}{2x} }} \Bigg) + \cos^{-1} \Bigg( \frac{1-x^2}{1+x^2} \Bigg) \Bigg\}$

$\displaystyle = \sin \Bigg\{ \sin^{-1} \Bigg( \frac{1-x^2}{1+x^2} \Bigg) + \cos^{-1} \Bigg( \frac{1-x^2}{1+x^2} \Bigg) \Bigg\}$

$\displaystyle \Big[ \because \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \Big]$

$\displaystyle = \sin \frac{\pi}{2} = 1 = \text{RHS }$

$\\$

$\displaystyle \text{Question 5: If:} \sin^{-1} \frac{2a}{1+a^2} + \sin^{-1} \frac{2b}{1+b^2} = 2\tan^{-1} x \text{ then prove that } \\ x = \frac{a+b}{1-ab}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } a = \tan z, \ \ \ b = \tan y$

$\displaystyle \text{Therefore, } \sin^{-1} \frac{2a}{1+a^2} + \sin^{-1} \frac{2b}{1+b^2} = 2\tan^{-1} x$

$\displaystyle \Rightarrow \sin^{-1} \frac{2\tan z}{1+\tan^2 z} + \sin^{-1} \frac{2\tan y}{1+\tan^2 y} = 2\tan^{-1} x$

$\displaystyle \Rightarrow \sin^{-1} (\sin 2z) + \sin^{-1} (\sin 2y) = 2 \tan^{-1} x$

$\displaystyle \Rightarrow z + y = \tan^{-1} x$

$\displaystyle \Rightarrow \tan^{-1} a + \tan^{-1} b = \tan^{-1} x$

$\displaystyle \Rightarrow \tan^{-1} \frac{a+b}{1-ab} = \tan^{-1} x$

$\displaystyle \Rightarrow \frac{a+b}{1-ab} = x$

$\\$

$\displaystyle \text{Question 6: Show that:} 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2} \text{ is a constant for } x \geq 1, \\ \text{find that constant}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Given, } 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2}$

$\displaystyle \text{(i) For } x > 1$

$\displaystyle = 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2}$

$\displaystyle = \pi - \sin^{-1} \frac{2x}{1+x^2} + \sin^{-1} \frac{2x}{1+x^2}$

$\displaystyle [\because 2 \tan^{-1} x = \pi - \sin^{-1} \Big( \frac{2x}{1+x^2} \Big), x > 1 \Big]$

$\displaystyle = \pi$

$\displaystyle \text{(i) For } x = 1$

$\displaystyle = 2 \tan^{-1} x + \sin^{-1} \frac{2x}{1+x^2}$

$\displaystyle = 2 \tan^{-1} 1 + \sin^{-1} \frac{2 \times 1}{1+1^2}$

$\displaystyle = 2 \tan^{-1} 1 + \sin^{-1} 1$

$\displaystyle = 2 \times \frac{\pi}{4} + \frac{\pi}{2}$

$\displaystyle = \frac{\pi}{2}+ \frac{\pi}{2} = \pi$

$\\$

$\displaystyle \text{Question 7: Find the value of each of the following:}$

$\displaystyle \text{(i)} \tan^{-1} \Bigg\{ 2 \cos \Bigg( 2 \sin^{-1} \frac{1}{2} \Bigg) \Bigg\}$

$\displaystyle \text{(ii)} \cos ( \sec^{-1} + \mathrm{cosec}^{-1} x ) , |x| \geq 1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i)}$

$\displaystyle \text{Let } \sin^{-1} \frac{1}{2} = y \Rightarrow \sin y = \frac{1}{2}$

$\displaystyle \text{Therefore, } \tan^{-1} \Bigg\{ 2 \cos \Bigg( 2 \sin^{-1} \frac{1}{2} \Bigg) \Bigg\}$

$\displaystyle = \tan^{-1} ( 2 \cos 2 y)$

$\displaystyle [\because \cos 2x = 1 - 2 \sin^2 x ]$

$\displaystyle = \tan^{-1} [ 2 ( 1 - 2 \sin^2 y) ]$

$\displaystyle = \tan^{-1} \Big[ 2 ( 1 - 2 \times \frac{1}{4}) \Big]$

$\displaystyle = \tan^{-1} (2 \times \frac{1}{2})$

$\displaystyle = \tan^{-1} 1$

$\displaystyle = \frac{\pi}{4}$

$\displaystyle \text{(ii)}$

$\displaystyle \text{Given, } \cos ( \sec^{-1} + \mathrm{cosec}^{-1} x ) = \cos \frac{\pi}{2} = 0$

$\displaystyle \Big[ \because \sec^{-1} x + \mathrm{cosec}^{-1} x = \frac{\pi}{2} \Big]$

$\\$

$\displaystyle \text{Question 8: Solve the following equations for:} x$

$\displaystyle \text{(i) } \tan^{-1} \frac{1}{4} + 2\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{6} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \text{(ii) } 3\sin^{-1} \frac{2x}{1+x^2} - 4\cos^{-1} \frac{1-x^2}{1+x^2} +2 \tan^{-1} \frac{2x}{1-x^2} = \frac{\pi}{3}$

$\displaystyle \text{(iii) } \tan^{-1} \frac{2x}{1-x^2} +\cot^{-1} \frac{1-x^2}{2x} = \frac{2\pi}{3} , x> 0 {\hspace{5.0cm} \text{[CBSE 2010]} }$

$\displaystyle \text{(iv) } 2 \tan^{-1} (\sin x) = \tan^{-1} (2 \sin x), x \neq \frac{\pi}{2}$

$\displaystyle \text{(v) } \cos^{-1} \frac{x^2 -1 }{x^2+1} +\frac{1}{2} \tan^{-1} \frac{2x}{1-x^2} = \frac{2\pi}{3} {\hspace{6.0cm} \text{[CBSE 2012]} }$

$\displaystyle \text{(vi) } \tan^{-1} \frac{x-2 }{x-1} +\tan^{-1} \frac{x+2}{x+1} = \frac{\pi}{4} {\hspace{7.0cm} \text{[CBSE 2016]} }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \text{LHS } = \tan^{-1} \frac{1}{4} + 2\tan^{-1} \frac{1}{5} + \tan^{-1} \frac{1}{6} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{2 \times \frac{1}{5}}{1 - (\frac{1}{5})^2} + \tan^{-1} \frac{1}{6} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{1}{4} + \tan^{-1} \frac{5}{12} + \tan^{-1} \frac{1}{6} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1}\Bigg\{ \frac{\frac{1}{4}+\frac{5}{12}}{1 -\frac{1}{4} \cdot\frac{5}{12} } \Bigg\} + \tan^{-1} \frac{1}{6} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{32}{43} + \tan^{-1} \frac{1}{6} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1}\Bigg\{ \frac{\frac{32}{43}+\frac{1}{6}}{1 -\frac{32}{43} \cdot\frac{1}{6} } \Bigg\} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{235}{226} + \tan^{-1} \frac{1}{x} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1}\Bigg\{ \frac{\frac{235}{226}+\frac{1}{x}}{1 -\frac{235}{226} \cdot\frac{1}{x} } \Bigg\} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{235x+226}{226x-235 } = \tan \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{235x+226}{226x-235 } = 1$

$\displaystyle \Rightarrow 235x+226 = 226x-235$

$\displaystyle \Rightarrow 9x = -461$

$\displaystyle \Rightarrow x = -\frac{461}{9}$

$\displaystyle \text{(ii) }$

$\displaystyle 3\sin^{-1} \frac{2x}{1+x^2} - 4\cos^{-1} \frac{1-x^2}{1+x^2} +2 \tan^{-1} \frac{2x}{1-x^2} = \frac{\pi}{3}$

$\displaystyle \Rightarrow 6 \tan^{-1} x - 8 \tan^{-1}x + 4 \tan^{-1} x = \frac{\pi}{3}$

$\displaystyle \Big[ \because 2 \tan^{-1} = \sin^{-1} \Big( \frac{2x}{1+x^2} \Big) \ \ \ \ \ \text{and } \ \ \ \ \ \ 2 \tan^{-1} = \cos^{-1} \Big( \frac{1-x^2}{1+x^2} \Big) \Big]$

$\displaystyle \Rightarrow 2 \tan^{-1} x = \frac{\pi}{3}$

$\displaystyle \Rightarrow \tan^{-1} x = \frac{\pi}{6}$

$\displaystyle \Rightarrow x = \tan \frac{\pi}{6}$

$\displaystyle \Rightarrow x = \frac{1}{\sqrt{3}}$

$\displaystyle \text{(iii) }$

$\displaystyle \text{Given, }\tan^{-1} \frac{2x}{1-x^2} +\cot^{-1} \frac{1-x^2}{2x} = \frac{2\pi}{3}$

$\displaystyle \Rightarrow \tan^{-1} \frac{2x}{1-x^2} +\tan^{-1} \frac{2x}{1-x^2} = \frac{2\pi}{3}$

$\displaystyle \Rightarrow 2\tan^{-1} \frac{2x}{1-x^2}= \frac{2\pi}{3}$

$\displaystyle \Big[ \because \cot^{-1} = \tan^{-1} \frac{1}{x} \Big]$

$\displaystyle \Rightarrow \tan^{-1} \frac{2x}{1-x^2}= \frac{\pi}{3}$

$\displaystyle \Big[ \because 2 \tan^{-1} x = \tan^{-1} \frac{2x}{1-x^2} \Big]$

$\displaystyle \Rightarrow 2\tan^{-1} x= \frac{\pi}{3}$

$\displaystyle \Rightarrow x= \tan\frac{\pi}{6}$

$\displaystyle \Rightarrow x= \frac{1}{\sqrt{3}}$

$\displaystyle \text{(iv) }$

$\displaystyle 2 \tan^{-1} (\sin x) = \tan^{-1} (2 \sin x), x \neq \frac{\pi}{2}$

$\displaystyle \Big[ \because 2 \tan^{-1} = \tan^{-1} \Big( \frac{2x}{1-x^2} \Big) \Big]$

$\displaystyle = \tan^{-1} \Big( \frac{2 \sin x}{1 - \sin^2 - x } \Big) = \tan^{-1} (2 \sin x)$

$\displaystyle \Rightarrow \frac{2 \sin x}{1 - \sin^2 - x } = 2 \sin x$

$\displaystyle \Rightarrow 2 \sin x = 2 \sin x - 2 \sin^3 x$

$\displaystyle \Rightarrow \sin^3 x = 0$

$\displaystyle \Rightarrow \sin x = 0$

$\displaystyle \Rightarrow x = 0$

$\displaystyle \text{(v) }$

$\displaystyle \text{Given, } \cos^{-1} \frac{x^2 -1 }{x^2+1} +\frac{1}{2} \tan^{-1} \frac{2x}{1-x^2} = \frac{2\pi}{3}$

$\displaystyle \Rightarrow 2 \tan^{-1} x +\frac{1}{2} \times 2 \tan^{-1} x = \frac{2\pi}{3}$

$\displaystyle \Rightarrow 3 \tan^{-1} x = \frac{2\pi}{3}$

$\displaystyle \Rightarrow x = \tan \frac{2 \pi}{9}$

$\displaystyle \text{(vi) }$

$\displaystyle \tan^{-1} \frac{x-2 }{x-1} +\tan^{-1} \frac{x+2}{x+1} = \frac{\pi}{4}$

$\displaystyle \Rightarrow \tan^{-1} \frac{x-2 }{x-1} +\tan^{-1} \frac{x+2}{x+1} = \tan^{-1} 1$

$\displaystyle \Rightarrow \tan^{-1} \frac{x-2 }{x-1} = \tan^{-1} 1 - \tan^{-1} \frac{x+2}{x+1}$

$\displaystyle \Rightarrow \tan^{-1} \frac{x-2 }{x-1} = \tan^{-1} \Bigg\{ \frac{1 - \frac{x+2}{x+1} }{1 + \frac{x+2}{x+1}} \Bigg\}$

$\displaystyle \Rightarrow \tan^{-1} \frac{x-2 }{x-1} = \tan^{-1} \Bigg\{ \frac{x+1 - x - 2}{x+1 + x + 2} \Bigg\}$

$\displaystyle \Rightarrow \tan^{-1} \frac{x-2 }{x-1} = \tan^{-1} \Bigg\{ \frac{-1}{2x+3} \Bigg\}$

$\displaystyle \Rightarrow \frac{x-2 }{x-1} = \frac{-1}{2x+3}$

$\displaystyle \Rightarrow 2x^2 + 3x - 4x - 6 = - x + 1$

$\displaystyle \Rightarrow 2x^2 = 7$

$\displaystyle \Rightarrow x = \pm \sqrt{\frac{7}{2}}$

$\\$

$\displaystyle \text{Question 14: Prove that: } 2 \tan^{-1} \Bigg( \sqrt{ \frac{a-b}{a+b}} \tan \frac{\theta}{2} \Bigg) = \cos^{-1} \Bigg(\frac{a \cos \theta + b}{a + b \cos \theta} \Bigg)$

$\displaystyle \text{Answer:}$

$\displaystyle \text{LHS } = 2 \tan^{-1} \Bigg( \sqrt{ \frac{a-b}{a+b}} \tan \frac{\theta}{2} \Bigg)$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{1 - \Big( \sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2} \Big)^2 }{1+\Big( \sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2} \Big)^2 } \Bigg\}$

$\displaystyle \Big[ \because 2 \tan^{-1} x = \cos^{-1} \frac{1-x^2}{1+x^2} \Big]$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{1 - \frac{a-b}{a+b} \tan^2 \frac{\theta}{2} }{1+ \frac{a-b}{a+b} \tan^2 \frac{\theta}{2} } \Bigg\}$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{a+b - (a-b) \tan^2 \frac{\theta}{2} }{a+b + (a-b) \tan^2 \frac{\theta}{2} } \Bigg\}$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{a+b - a\tan^2 \frac{\theta}{2} + b \tan^2 \frac{\theta}{2} }{a+b + a\tan^2 \frac{\theta}{2} - b \tan^2 \frac{\theta}{2} } \Bigg\}$

$\displaystyle \Big[ \text{Dividing numerator and denominator by } 1 +\tan^2 \frac{\theta}{2} \Big]$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{a \Bigg( \frac{1 - \tan^2 \frac{\theta}{2}}{1 +\tan^2 \frac{\theta}{2} } \Bigg)+ b\Bigg( \frac{1 + \tan^2 \frac{\theta}{2}}{1 +\tan^2 \frac{\theta}{2} } \Bigg) }{a\Bigg( \frac{1 + \tan^2 \frac{\theta}{2}}{1 +\tan^2 \frac{\theta}{2} } \Bigg) + b\Bigg( \frac{1 - \tan^2 \frac{\theta}{2}}{1 +\tan^2 \frac{\theta}{2} } \Bigg) } \Bigg\}$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{a \Bigg( \frac{1 - \tan^2 \frac{\theta}{2}}{1 +\tan^2 \frac{\theta}{2} } \Bigg)+ b }{a + b\Bigg( \frac{1 - \tan^2 \frac{\theta}{2}}{1 +\tan^2 \frac{\theta}{2} } \Bigg) } \Bigg\}$

$\displaystyle = \cos^{-1} \Bigg\{ \frac{a \cos \theta+ b }{a + b\cos \theta } \Bigg\} = \text{RHS }$

$\\$

$\displaystyle \text{Question 15: Prove that:} \tan^{-1} \frac{2ab}{a^2 - b^2} + \tan^{-1} \frac{2xy}{x^2 - y^2} = \tan^{-1} \frac{2 \alpha \beta}{\alpha^2 - \beta^2} \\ \\ \text{ where } \alpha = a x - by \text{ and } \beta = ay + bx$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We know, } \tan^{-1} x + \tan^{-1} y = \tan^{-1} \Big( \frac{x-y}{1+xy} \Big), xy>1$

$\displaystyle \therefore \tan^{-1} \frac{2ab}{a^2 - b^2} + \tan^{-1} \frac{2xy}{x^2 - y^2}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{\frac{2ab}{a^2 - b^2} + \frac{2xy}{x^2 - y^2}}{1 - \frac{2ab}{a^2 - b^2} \cdot \frac{2xy}{x^2 - y^2}} \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{ \frac{2(abx^2 - aby^2 + xya^2 - xyb^2)}{(a^2 - b^2)(x^2 - y^2)} }{ \frac{(a^2x^2-a^2y^2-x^2b^2+y^2b^2-4abxy)}{(a^2 - b^2)(x^2 - y^2)} } \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2(abx^2 - aby^2 + xya^2 - xyb^2)}{(a^2x^2-a^2y^2-x^2b^2+y^2b^2-4abxy)} \Bigg\}$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2(ax-by)(ay+bx)}{(ax-by)^2 - (ay+bx)^2} \Bigg\}$

$\displaystyle \text{Since, } \alpha = a x - by \text{ and } \beta = ay + bx$

$\displaystyle = \tan^{-1} \Bigg\{ \frac{2\alpha \beta }{\alpha ^2 - \beta^2} \Bigg\}$

$\\$

$\displaystyle \text{Question 16: For any:} a, b, x, y, > 0 \text{ prove that: } \\ \\ \frac{2}{3} \tan^{-1} \Bigg( \frac{3ab^2- a^3}{b^3 - 3a^2b} \Bigg) + \frac{2}{3} \tan^{-1} \Bigg( \frac{3xy^2- x^3}{y^3 - 3x^2y} \Bigg) = \tan^{-1} \Bigg( \frac{2 \alpha \beta}{\alpha^2 - \beta^2}\Bigg) \\ \\ \text{ where } \alpha = - ax + by , \beta = bx + ay$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let }a = b \tan m \text{ and } x = y \tan n$

$\displaystyle \text{Given, }\frac{2}{3} \tan^{-1} \Bigg( \frac{3ab^2- a^3}{b^3 - 3a^2b} \Bigg) + \frac{2}{3} \tan^{-1} \Bigg( \frac{3xy^2- x^3}{y^3 - 3x^2y} \Bigg)$

$\displaystyle = \frac{2}{3} \tan^{-1} \Bigg( \frac{3b^3 \tan m- b^3 \tan^3 m}{b^3 - 3b^3 \tan^2 m} \Bigg) + \frac{2}{3} \tan^{-1} \Bigg( \frac{3y^3 \tan n- y^3 \tan^3 n}{y^3 - 3y^3 \tan^2 n} \Bigg)$

$\displaystyle = \frac{2}{3} \tan^{-1} \Bigg( \frac{3 \tan m- \tan^3 m}{1 - 3 \tan^2 m} \Bigg) + \frac{2}{3} \tan^{-1} \Bigg( \frac{3 \tan n- \tan^3 n}{1 - 3 \tan^2 n} \Bigg)$

$\displaystyle = \frac{2}{3} \tan^{-1} ( \tan 3m ) + \frac{2}{3} \tan^{-1} ( \tan 3n )$

$\displaystyle \Big[ \because \tan 3x = \frac{3 \tan x- \tan^3 x}{1 - 3 \tan^2 x} \Big]$

$\displaystyle = \frac{2}{3} 3m + \frac{2}{3} 3n$

$\displaystyle = 2m+2n$

$\displaystyle = 2 \Big(\tan^{-1} \frac{a}{b} + \tan^{-1} \frac{x}{y} \Big)$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{ \frac{a}{b} + \frac{x}{y} }{1 - \frac{a}{b} \cdot \frac{x}{y} } \Bigg\}$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{ ay+bx }{by-ax } \Bigg\}$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{ 2 \times \frac{ ay+bx }{by-ax } }{1 - \Big( \frac{ ay+bx }{by-ax } \Big)^2 } \Bigg\}$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{2(ay+bx) (by-ax)}{ (by-ax)^2 - (ay+bx)^2 } \Bigg\}$

$\displaystyle = 2 \tan^{-1} \Bigg\{ \frac{2 \alpha \beta }{ \alpha^2 - \beta^2 } \Bigg\}$

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Class 12: Inverse Trigonometric Functions – Exercise 4.14

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